Find the negative root of the equation x^3 - 2x + 5 =0
Find a root of an equation f(x)=x3-2x+5 using Bisection method
Solution:
Here x3-2x+5=0
Let f(x)=x3-2x+5
Here
x 0 -1 -2 -3 -4
f(x) 5 6 1 -16 -51
1st iteration :
Here f(-3)=-16<0 and f(-2)=1>0
∴ Now, Root lies between -3 and -2
x0=-3+(-2)2=-2.5
f(x0)=f(-2.5)=(-2.5)3-2⋅(-2.5)+5=-5.625<0
2nd iteration :
Here f(-2.5)=-5.625<0 and f(-2)=1>0
∴ Now, Root lies between -2.5 and -2
x1=-2.5+(-2)2=-2.25
f(x1)=f(-2.25)=(-2.25)3-2⋅(-2.25)+5=-1.8906<0
3rd iteration :
Here f(-2.25)=-1.8906<0 and f(-2)=1>0
∴ Now, Root lies between -2.25 and -2
x2=-2.25+(-2)2=-2.125
f(x2)=f(-2.125)=(-2.125)3-2⋅(-2.125)+5=-0.3457<0
4th iteration :
Here f(-2.125)=-0.3457<0 and f(-2)=1>0
∴ Now, Root lies between -2.125 and -2
x3=-2.125+(-2)2=-2.0625
f(x3)=f(-2.0625)=(-2.0625)3-2⋅(-2.0625)+5=0.3513>0
5th iteration :
Here f(-2.125)=-0.3457<0 and f(-2.0625)=0.3513>0
∴ Now, Root lies between -2.125 and -2.0625
x4=-2.125+(-2.0625)2=-2.0938
f(x4)=f(-2.0938)=(-2.0938)3-2⋅(-2.0938)+5=0.0089>0
6th iteration :
Here f(-2.125)=-0.3457<0 and f(-2.0938)=0.0089>0
∴ Now, Root lies between -2.125 and -2.0938
x5=-2.125+(-2.0938)2=-2.1094
f(x5)=f(-2.1094)=(-2.1094)3-2⋅(-2.1094)+5=-0.1668<0
7th iteration :
Here f(-2.1094)=-0.1668<0 and f(-2.0938)=0.0089>0
∴ Now, Root lies between -2.1094 and -2.0938
x6=-2.1094+(-2.0938)2=-2.1016
f(x6)=f(-2.1016)=(-2.1016)3-2⋅(-2.1016)+5=-0.0786<0
8th iteration :
Here f(-2.1016)=-0.0786<0 and f(-2.0938)=0.0089>0
∴ Now, Root lies between -2.1016 and -2.0938
x7=-2.1016+(-2.0938)2=-2.0977
f(x7)=f(-2.0977)=(-2.0977)3-2⋅(-2.0977)+5=-0.0347<0
9th iteration :
Here f(-2.0977)=-0.0347<0 and f(-2.0938)=0.0089>0
∴ Now, Root lies between -2.0977 and -2.0938
x8=-2.0977+(-2.0938)2=-2.0957
f(x8)=f(-2.0957)=(-2.0957)3-2⋅(-2.0957)+5=-0.0129<0
10th iteration :
Here f(-2.0957)=-0.0129<0 and f(-2.0938)=0.0089>0
∴ Now, Root lies between -2.0957 and -2.0938
x9=-2.0957+(-2.0938)2=-2.0947
the negative root of the equation =-2.0947
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